AgNO_{3} + KCl → AgCl + KNO_{3}

**$\mathbf{0}\mathbf{.}\mathbf{780}\mathbf{}\overline{)\mathbf{mg}\mathbf{}\mathbf{KCl}}\mathbf{}\mathbf{\times}\frac{\mathbf{1}\mathbf{\times}{\mathbf{10}}^{\mathbf{-}\mathbf{3}}\mathbf{}\overline{)\mathbf{g}\mathbf{}\mathbf{KCl}}}{\mathbf{1}\mathbf{}\overline{)\mathbf{mg}\mathbf{}\mathbf{KCl}}}\mathbf{\times}\frac{\mathbf{1}\mathbf{}\overline{)\mathbf{mol}\mathbf{}\mathbf{KCl}}}{\mathbf{74}\mathbf{.}\mathbf{55}\mathbf{}\overline{)\mathbf{g}\mathbf{}\mathbf{KCl}}}\mathbf{\times}\frac{\mathbf{1}\mathbf{}\mathbf{mol}\mathbf{}{\mathbf{AgNO}}_{\mathbf{3}}}{\mathbf{1}\mathbf{}\overline{)\mathbf{mol}\mathbf{}\mathbf{KCl}}}$**

** = 1.05x10 ^{-5} mol AgNO_{3}**

You may want to reference (Pages 149 - 153) Section 4.6 while completing this problem.

If 26.2 mL of AgNO_{3} is needed to precipitate all the Cl^{-} ions in a 0.780-mg sample of KCl (forming AgCl), what is the molarity of the AgNO_{3} solution?

Frequently Asked Questions

What scientific concept do you need to know in order to solve this problem?

Our tutors have indicated that to solve this problem you will need to apply the Solution Stoichiometry concept. You can view video lessons to learn Solution Stoichiometry. Or if you need more Solution Stoichiometry practice, you can also practice Solution Stoichiometry practice problems.